--%>

Schrodinger equation with particle in a box problem.

Three dimensional applications of the Schrodinger equation are introduced by the particle-in-a-box problem.

So far only a one-dimensional problem has been solved by application of the Schrodinger equation. Now the allowed energies and the probability functions for a particle that is free to move in three dimensions are deduced. A molecule of a gas enclosed in a cubic container provides a specific example that is dealt with in the section after the general procedure has been developed.

For any three-dimensional problem, the potential energy is, general, a function of three coordinates. For a cubic potential box, the Cartesian coordinates are convenient. The differential equation that must be solved is now the Schrodinger equation in three dimensions.

1310_Particle in a box.png 

For a "cubic box," the potential function can be expressed in terms of separate x, y, and z components,
98_Particle in a box1.png 

Each of the potential function components for a "particle-in-a-box" is like the one-dimensional potential for a "particle-on-a-line".

For three-dimensional systems, the solution function ψ depends on the three coordinates necessary to locate a point in space. It is often profitable to try to separate such systems into parts, with each part involving only one coordinate. On the basis we try the substitution

ψ (x, y, z) = Ø(x)Ø(y)Ø(z)

Substitution of (2) from (1) gives

1350_Particle in a box2.png 

Division by Ø(x)Ø(y)Ø(z) gives

1180_Particle in a box3.png 

For the equation to be satisfied for all values of x, y and z, each term on the left must equal a component of ε, and we can write

ε = εx + εy + εz

The Schrodinger equation can then be broken down into three identical equations of the type

1394_Particle in a box4.png 

Or

578_Particle in a box6.png 

These equations are identical to that written for one-dimensional problem. The solution to the three-dimensional cubic-box problem is therefore

ψ =  Ø(x)Ø(y)Ø(z)

With

1809_Particle in a box7.png

   Related Questions in Chemistry

  • Q : What is ortho effect? Orthosubstituted

    Orthosubstituted anilines are generally weaker bases than aniline irrespective of the electron releasing or electron withdrawing nature of the substituent. This is known as ortho effect and may probably be due to combined electronic and steric factors.The overall basic strength of ort

  • Q : Dipole attractions-London dispersion

    Describe how dipole attractions, London dispersion forces and the hydrogen bonding identical?

  • Q : Solubility are halides are halogens

    are halides are halogens more soluble? why?

  • Q : Describe the function of the

    Briefly describe the function of the monosaccharide?

  • Q : Vapour pressure of volatile substance

    Provide solution of this question. According to Raoult's law the relative lowering of vapour pressure of a solution of volatile substance is equal to: (a) Mole fraction of the solvent (b) Mole fraction of the solute (c) Weight percentage of a solute (d) Weight perc

  • Q : Number of mlecules in methane Can

    Can someone please help me in getting through this problem. The total number of molecules in 16 gm of methane will be: (i) 3.1 x 1023 (ii) 6.02 x 1023 (iii) 16/6.02 x 1023 (iv) 16/3.0 x 1023

  • Q : What are heterogenous catalysis? Give

    When the catalyst exists in a different phase than that of reactants, it is said to be heterogeneous catalyst, and the catalysis is called heterogeneous catalysis. For example, SO2 can be oxidized to SO3

  • Q : Explain equilibrium and molecular

    The equilibrium constant can be treated as a particular type of molecular distribution. Consider the simplest gas-phase reaction, one in which molecules of A are converted to molecules of B. the reaction, described by the equation

    Q : Eutectic Formation In some two

    In some two component, solid liquid systems, a eutectic mixture forms.Consider, now a two component system at some fixed pressure, where the temperature range treated is such as to include formation of one or more solid phases. A simple behavior is shown b

  • Q : Molality of glucose Help me to go

    Help me to go through this problem. Molecular weight of glucose is 180. A solution of glucose which contains 18 gms per litre is : (a) 2 molal (b) 1 molal (c) 0.1 molal (d)18 molal